hdu 1402 FFT（模板）

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A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16111 Accepted Submission(s): 3261

Problem Description

Calculate A * B.

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output

For each case, output A * B in one line.

Sample Input

Sample Output

Author

DOOM III

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题意：求高精度a*b --代码参考kuangbin大神

思路：

通过FFT我们可以快速求出多项式的卷积，从而解决数相乘。

求卷积大致如下图，至于FFT具体原理看不太懂- -

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>using namespace std;typedef long long ll;typedef long double ld;const ld eps=1e-10;const int inf = 0x3f3f3f;const int MOD = 1e9+7;const double PI = acos(-1.0);struct Complex{    double x,y;    Complex(double _x = 0.0,double _y = 0.0)    {        x = _x;        y = _y;    }    Complex operator-(const Complex &b)const    {        return Complex(x-b.x,y-b.y);    }    Complex operator+(const Complex &b)const    {        return Complex(x+b.x,y+b.y);    }    Complex operator*(const Complex &b)const    {        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);    }};void change(Complex y[],int len){    int i,j,k;    for(i = 1,j = len/2; i < len-1; i++)    {        if(i < j) swap(y[i],y[j]);        k = len/2;        while(j >= k)        {            j-=k;            k/=2;        }        if(j < k) j+=k;    }}void fft(Complex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));        for(int j = 0; j < len; j+=h)        {            Complex w(1,0);            for(int k = j; k < j+h/2; k++)            {                Complex u = y[k];                Complex t = w*y[k+h/2];                y[k] = u+ t;                y[k+h/2] = u-t;                w = w*wn;            }        }    }    if(on == -1)    {        for(int i = 0; i < len; i++)            y[i].x /= len;    }}const int maxn = 200100;Complex x1[maxn],x2[maxn];char str1[maxn],str2[maxn];int sum[maxn];int main(){    while(scanf("%s%s",str1,str2) != EOF)    {        int len1 = strlen(str1);        int len2 = strlen(str2);        int len = 1;        while(len < len1*2 || len < len2*2) len <<= 1;        for(int i = 0; i < len1; i++)            x1[i] = Complex(str1[len1-i-1]-'0',0);        for(int i = len1; i < len; i++)            x1[i] = Complex(0,0);        for(int i = 0; i < len2; i++)            x2[i] = Complex(str2[len2-1-i]-'0',0);        for(int i = len2; i < len; i++)            x2[i] = Complex(0,0);        fft(x1,len,1);        fft(x2,len,1);        for(int i = 0; i < len; i++)        {            x1[i] =x1[i]*x2[i];            //cout << x1[i].x << " "<< x1[i].y <<endl;        }        fft(x1,len,-1);        for(int i = 0;i < len;i++){            sum[i] = (int)(x1[i].x+0.5);            //cout << sum[i] << endl;        }        for(int i = 0; i < len; i++)        {            sum[i+1] += sum[i]/10;            sum[i] %= 10;        }        len= len1+len2-1;        while(sum[len] <= 0 && len > 0)            len--;        for(int i = len; i >= 0; i--)            printf("%c",sum[i]+'0');        printf("\n");    }    return 0;}

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